How to derive matrix form of the Bellman operators?
There's not much to derive here it's simply a definition of Bellman operator, it comes from Bellman equation. If you're wondering why
\begin{equation}
Q^{\pi} = (I - \gamma P^{\pi})^{-1}r \tag{1}
\end{equation}
they state that $Q^{\pi}$ is a fixed point which means if you apply Bellman operator to it you get the same value
\begin{equation}
T^{\pi}(Q^{\pi}) = Q^{\pi}
\end{equation}
You can easily check that since from $(1)$
\begin{equation}
r = (I-\gamma P^{\pi})Q^{\pi}
\end{equation}
if you plug it in definition of Bellman operator you get
\begin{align}
T^{\pi}(Q^{\pi}) &amp;= r + \gamma P^{\pi} Q^{\pi}\\
&amp;= (I - \gamma P^{\pi})Q^{\pi}+  \gamma P^{\pi} Q^{\pi}\\
&amp;= Q^{\pi}
\end{align}